Question

For which 2 of the basic trig identities is one side of the equation defined at \(\theta=0\) and the other side isn't?

Answer 1

@jim_thompson5910

Answer 2

anyone??

Answer 3

for it to be an identity, it would have to be an equation that is defined for all allowed input values

so to say that it's defined on one side, but not on the other, doesn't make sense because that would invalidate the fact it's an identity

Answer 4

but if you're asking which trig functions are undefined at theta = 0, they are

csc(theta) and cot(theta)

because

csc(x) = 1/sin(x)

cot(x) = cos(x)/sin(x)

and sin(x) = 0 when x = 0

Answer 5

i dont know but the exact question from my book is:

For exactly teo of the basic identities, one side of the equation is defined at \(\theta=0\) and the other side is not. Which two?

Answer 6

??

Answer 7

oh i see what it means, if you check out this page (which I'm assuming what your book looks like)

http://fhs.fusdaz.org/download.axd?file=f908ec34-dd0e-44c2-b0aa-e39839832177&dnldType=Resource

they list the basic identities of which you use to answer this question

Answer 8

what I'm referring to is this

Answer 9

yes i have that but i cant seem to figure it out

Answer 10

so which one of these identities have one side defined at theta = 0, but theta = 0 is undefined on the other side?

Answer 11

okay i think its \[sin\theta={1\over csc\theta}\]and\[tan\theta={1\over cot\theta}\]

Answer 12

am i right?

Answer 13

correct, sine is defined at theta = 0, but csc is not

same with tan and cot

Answer 14

okay one more question

Answer 15

how do you do:\[{tan^2x\over sec~x+1}\]

Answer 16

ir says to write the expression as an algebraic expression of a single trigonometric function....and it says the answer is \(sec~x+1\)

Answer 17

from this page,

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

1 + tan^2x = sec^2x

tan^2x = sec^2x - 1

tan^2x = (secx + 1)(secx - 1)

Answer 18

then shouldnt the answer be secx-1 not +1?

Answer 19

there must be a typo then

Answer 20

in \[sin~x~tan^2x-sin~x=0\]how do you factor out sin x?

Answer 21

like this

sin(x)*tan^2(x) - sin(x) = 0

sin(x) [ tan^2(x) - 1 ] = 0

Answer 22

im stupid..

Answer 23

no you're not, just a silly mistake

Answer 24

okay so how do you go on from there:\[sin~x=0~~~~;~~~~tan^2x=0\]it asks for what values of x sin x would equal 0...i dont get it

Answer 25

tan^2(x)-1=0* sorry

Answer 26

when is sin(x) = 0 true

Answer 27

yea exactly...i dont get how you would find that

Answer 28

wait is it 1 and -1?

Answer 29

do you have the unit circle with you?

Answer 30

i know it, yea

Answer 31

sin x = 0 when x=+/-1

Answer 32

right?

Answer 33

no

Answer 34

use this

http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png

Answer 35

locate the angles when the y coordinate is 0

Answer 36

pi/2 and 3pi/2?

Answer 37

yes??

Answer 38

close, you're seeing the x coordinate as zero there though

it's actually 0 and pi

Answer 39

oh right

Answer 40

and what about the other one?\[tan^2x-1=0~~\implies~~~tan~x=\pm~\underline{~~~~~~~~}?\]

Answer 41

\[\pm1\]right?

Answer 42

tan^2x-1 = 0

(tanx - 1)(tanx + 1) = 0

tanx - 1 = 0 or tanx + 1 = 0

tanx = 1 or tanx = -1

so yes, plus/minus 1

Answer 43

okay and can you check y work for this one:\[2~sin^2x+3~sin~x=2\]\[2~sin^2x+3~sin~x-2=0\]\[(2~sin~x-1)(sin~x+2)=0\]\[2~sin~x-1=0~~~~~~~~~~~~~~sin~x+2=0\]\[sin~x={1\over2};{-2}\]yea?

Answer 44

you got it, just solve each for x from here

Answer 45

so that would be pi/6 and...whats the other one?

Answer 46

isnt sin x=-2 a no solution since sin x can only be between -1 and 1 ??

Answer 47

correct, there are no solutions to sin(x) = -2

Answer 48

sin(x) = 1/2 has the solutions

x = pi/6

and x = 5pi/6

the other solution is found by subtracting pi/6 from pi

pi - pi/6 = 6pi/6 - pi/6 = 5pi/6

Answer 49

this is assuming 0 < x < 2pi

Answer 50

wouldnt it be 13pi/6?

Answer 51

how are you getting that?

Answer 52

on my sheet it says to add 2pi to the solution to get all the values of x

Answer 53

oh gotcha

Answer 54

well that just means add multiples of 2pi, so

one is pi/6 + 2pi*k where k is any integer

another is 5pi/6 + 2pi*k where k is any integer

Answer 55

yea thats what it says

Answer 56

you just leave it like that though

Answer 57

okay and if you were to represent this expression involving only sines and cosines would this be it?\[{{(sec~y-tan~y)(sec~y+tan~y)}\over{sec~y}}~~~~=~~~~{cos~y}\]

Answer 58

yes??

Answer 59

one sec

Answer 60

k

Answer 61

sry got distracted, but yes that's correct

Answer 62

the numerator turns into sec^2 - tan^2, which just 1

Answer 63

lol no problem...cool :)

Answer 64

hey brb i have a few more...thank you sooo much so far

Answer 65

if tan x = (pi/2 - theta) = -5.32, find cot theta.

Answer 66

and it gives the blanks:\[Since~~tan~x=({\pi\over2}-\theta)=~\underline{~~~~~~~~~~}~~then~~cot\theta=\underline{~~~~~~~~~}\]

Answer 67

same page

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 68

look where it says "Co-Function Identities"

Answer 69

so is that\[tan~x=({\pi\over2}-\theta)=-5.32\]\[cot\theta=-{1\over5.32}\]yes?

Answer 70

no, tan(pi/2 - x) = cot(x) according to the page

Answer 71

ohhhh right sorry i didnt look at that ...so it would be cot theta=-5.32?

Answer 72

yep

Answer 73

take (sin x)^2 + (cox x)^2 = 1 and divide through by (cos x)^2, what do you get?
take (sin x)^2 + (cox x)^2 = 1 and divide through by (sin x)^2, what do you get?

i got (sec x)^2 and (csc x)^2

Answer 74

(sin x)^2 + (cox x)^2 = 1

turns into

tan^2 + 1 = 1/cos^2

after dividing everything by cos^2

which is really

tan^2 + 1 = sec^2 (something we've already seen and used before in this post)

-------------------------------------------------------

(sin x)^2 + (cox x)^2 = 1

turns into

1 + (cot x)^2 = 1/sin^2

after dividing everything by sin^2

which is really

1 + cot^2 = csc^2

these two identities are listed on that page I gave you

Answer 75

so its what i said?

Answer 76

yeah part of what you said is correct, but i think they wanted those identities

Answer 77

im a little confused with what you did up there...i cant find that anywhere on the page you gave

Answer 78

look for "Pythagorean Identities"

Answer 79

second group on that page

Answer 80

yea but none of those say what you said

Answer 81

i just left off the variable

Answer 82

okay can you explain it again a little simpler or something,idk i didnt get it the first time

Answer 83

(sin x)^2 + (cos x)^2 = 1

(sin x)^2/(cos x)^2 + (cos x)^2/(cos x)^2 = 1/(cos x)^2

(tan x)^2 + 1 = (sec x )^2

Answer 84

in that I'm dividing everything by (cos x)^2

Answer 85

ohh

Answer 86

yeah i just did it without the variables

Answer 87

okay so its {(tan x)^2 + 1 = (sec x )^2}/(cos x)^2

Answer 88

how are you getting that?

Answer 89

when it says to divide by (cos x)^2

Answer 90

yeah but when it turns into sec x, the cosine on the right side goes away

Answer 91

so it's just (tan x)^2 + 1 = (sec x )^2

Answer 92

oh so thats the answer?

Answer 93

yeah for that first part

Answer 94

and then the second one would be 1+ (cot x)^2 = (csc x)^2

Answer 95

yeah

Answer 96

okay yay!
tan theta = 3 and cos theta > 0

since (tan theta)^2 +1 = (sec theta)^2 then 9 + 1 = (sec theta)^2, 10= (sec theta)^2, sqrt10= sec theta
since cos theta > 0 then sec theta = -10
so cos theta = -1/10

yes?

Answer 97

that was the last one. is that right?